Probability Formulas

Introduction to Probability

Sample–Space: The set of all possible outcomes of an experiment is called the Sample–Space(S).
Event: A subset of sample−space is called an Event.
Complement Of An Event A: The set of all out comes which are in S but not in A is called the Complement Of An Event A Denoted By $\bar {A}$ Or A^c.
Compound Event: If A & B are two given events then A∩B is called Compound Event and is denoted by A∩B or AB or A & B.
Mutually Exclusive Events: Two events are said to be Mutually Exclusive Events (or disjoint or incompatible) if the occurence of one precludes (rules out) the simultaneous occurence of the other. If A & B are two mutually exclusive events then P (A & B) = 0.
Equally Likely Events: Events are said to be Equally Likely when each event is as likely to occur as any other event.
Exhaustive Events: Events A,B,C …….. L are said to be Exhaustive Events if no event outside this set can result as an outcome of an experiment. For example, if A & B are two events defined on a sample space S, then A & B are exhaustive ⇒ A∪B = S⇒ P (A∪B) = 1.
Classical Def. Of Probability: If n represents the total number of equally likely, mutually exclusive and exhaustive outcomes of an experiment and m of them are favourable to the happening of the event A, then the probability of happening of the event A is given by P(A) = m/n.
Note:
(1) 0 ≤ P(A) ≤ 1
(2) P(A) + P( $\bar {A}$ ) = 1, Where $\bar {A}$ = Not A .
(3) If x cases are favourable to $\bar {A}$ & y cases are favourable to A then P(A) = $\frac {x}{x+y}$ and P( $\bar {A}$ ) = $\frac {y}{x+y}$
We say that Odds In Favour Of A are x: y & odds against A are y : x

Comparative study of Equally likely , Mutually Exclusive and Exhaustive events.

Experiment	Events	E/L	M/E	Exhaustive
1. Throwing of a die	A: throwing an odd face { 1, 3, 5} B: throwing a composite face { 4, 6}	No	Yes	No
2. A ball is drawn from an urn containing 2W, 3R and 4G balls	E₁: getting a W ball E₂: getting a R ball E₃: getting a G ball	No	Yes	Yes
3. Throwing a pair of dice	A: throwing a doublet { 11, 22, 33, 44, 55, 66} B: throwing a total of 10 or more { 46, 64, 55, 56, 65, 66}	Yes	No	No
4. From a well shuffled pack of cards a card is drawn	E₁: getting a heart E₂: getting a spade E₃: getting a diamond E₄: getting a club	Yes	Yes	Yes
5. From a well shuffled pack of cards a card is drawn	A = getting a heart B = getting a face card	No	No	No

Results − 2

AUB = A+ B = A or B denotes occurence of at least A or B. For 2 events A & B : (See fig.1)

P(A∪B) = P(A) + P(B) − P(A∩B) = P(A. $\bar {B}$ ) + P( $\bar {A}$ .B) + P(A.B) = 1 − P( $\bar {A}.$ $\bar {B}$ )
Opposite of “atleast A or B” is Niether A nor B i.e. $\bar {A + B}$ = 1-(A or B) = $\bar {A}$ ∩ $\bar {B}$
Note that P(A+B) + P( $\overline{A} \cap \overline{B}$ ) = 1.
IfA & B are mutually exclusive then P(A∪B) = P(A) + P(B).
For any two events A & B, P(exactly one of A , B occurs)
= P (A ∩ $\bar {B}$ ) + P (B ∩ $\bar {A}$ ) = P (A) + P (B) − 2P (A ∩ B)
= P (A ∪ B) − P (A ∩ B) = P (A^c ∪ B^c ) − P(A^c ∩ B^c )
If A & B are any two events P(A∩B) = P(A).P(B/A) = P(B).P(A/B), Where P(B/A) means conditional
probability of B given A & P(A/B) means conditional probability ofA given B. (This can be easily seen
from the figure)
DE MORGAN’S LAW : − IfA & B are two subsets of a universal set U , then
- (A∪B)c = Ac∩Bc &
- (A∩B)c = Ac∪Bc
A ∪ (B∩C) = (A∪B) ∩ (A∪C) & A ∩ (B∪C) = (A∩B) ∪ (A∩C)

Result − 3

For any three events A,B and C we have (See Fig. 2)

P(A or B or C) = P(A) + P(B) + P(C) − P(A∩B) − P(B∩C)− P(C∩A) + P(A∩B∩C)
P (at least two of A,B,C occur) = P(B∩C) + P(C∩A) + P(A∩B) − 2P(A∩B∩C)
P(exactly two of A,B,C occur) = P(B∩C) + P(C∩A) + P(A∩B) − 3P(A∩B∩C)
P(exactly one of A,B,C occurs) = P(A) + P(B) + P(C) − 2P(B∩C) − 2P(C∩A) − 2P(A∩B)+3P(A∩B∩C)
Note: If three events A, B and C are pair wise mutually exclusive then they must be mutually exclusive.
i.e P(A∩B) = P(B∩C) = P(C∩A) = 0 ⇒ P(A∩B∩C) = 0. However the converse of this is not true.

Result − 4
Independent Events: Two events A & B are said to be independent if occurence or non occurence of one does not effect the probability of the occurence or non occurence of other.

If the occurence of one event affects the probability of the occurence of the other event then the events are said to be Dependent or Contingent. For two independent events A and B : P(A∩B) = P(A). P(B). Often this is taken as the definition of independent events.
Three events A , B & C are independent if & only if all the following conditions hold;
P(A∩B) = P(A) . P(B) ; P(B∩C) = P(B) . P(C)
P(C∩A) = P(C) . P(A) & P(A∩B∩C) = P(A) . P(B) . P(C)
i.e. they must be pairwise as well as mutually independent.
Similarly for n events A₁, A₂, A₃, …… A_n to be independent, the number of these conditions is equal to ⁿc₂ + ⁿc₃ + ….. + ⁿc_n = 2ⁿ − n − 1.
The probability of getting exactly r success in n independent trials is given by P(r) = ⁿC_rp^r q^n-r
where: p = probability of success in a single trial q = probability of failure in a single trial. note : p + q = 1
Note: Independent events are not in general mutually exclusive & vice versa. Mutually exclusiveness can be used when the events are taken from the same experiment & independence can be used when the events are taken from different experiments.

Result − 5:
Baye’s Theorem Or Total Probability Theorem:
If an event A can occur only with one of the n mutually exclusive and exhaustive events B₁, B₂,… B_n & the probabilities P(A/B₁), P(A/B₂) ……. P(A/B_n) are known then,

$P\left(B_{1} / A\right)=\frac{P\left(B_{i}\right) \cdot P\left(A / B_{i}\right)^{2}}{\sum_{i=1}^{n} P\left(B_{i}\right) \cdot P\left(A / B_{i}\right)}$
Proof:
The events A occurs with one of the n mutually exclusive & exhaustive events B₁, B₂,B₃,……..B_n; A = AB₁ + AB₂ + AB₃ + ……. + AB_n
$\mathrm{P}(\mathrm{A})=\mathrm{P}\left(\mathrm{AB}_{1}\right)+\mathrm{P}\left(\mathrm{AB}_{2}\right)+\ldots \ldots+\mathrm{P}\left(\mathrm{AB}_{\mathrm{r}}\right)=\sum_{i=1}^{\mathrm{n}} \mathrm{P}\left(\mathrm{AB}_{\mathrm{i}}\right)$
Note: A ≡ event what we have;
B₁≡ event what we want ;
B₂, B₃, ….B_n are alternative event .
Now, P(AB_i) = P(A) . P(B_i/A)
= P(B_i) . P(A/B_i)
$P\left(B_{1} / A\right)=\frac{P\left(B_{1}\right) \cdot P\left(A / B_{i}\right)}{P(A)}=\frac{P\left(B_{1}\right) \cdot P\left(A / B_{i}\right)}{\sum_{i=1}^{n} P\left(A B_{i}\right)}$
$\mathrm{P}\left(\mathrm{B}_{\mathrm{i}} / \mathrm{A}\right)=\frac{\mathrm{P}\left(\mathrm{B}_{\mathrm{i}}\right) \cdot \mathrm{P}\left(\mathrm{A} / \mathrm{B}_{\mathrm{i}}\right)}{\sum \mathrm{P}\left(\mathrm{B}_{\mathrm{i}}\right) \cdot \mathrm{P}\left(\mathrm{A} / \mathrm{B}_{\mathrm{i}}\right)}$

Result − 6
If p1 and p2 are the probabilities of speaking the truth of two indenpendent witnesses A and B thenP (their combined statement is true)
$=\frac{\mathrm{p}_{1} \mathrm{p}_{2}}{\mathrm{p}_{1} \mathrm{p}_{2}+\left(1-\mathrm{p}_{1}\right)\left(1-\mathrm{p}_{2}\right)}.$
In this case it has been assumed that we have no knowledge of the event except the statement made by A and B. However, if p is the probability of the happening of the event before their statement then P (their combined statement is true)
$=\frac{\mathrm{p} \mathrm{p}_{1} \mathrm{p}_{2}}{\mathrm{pp}_{1} \mathrm{p}_{2}+(1-\mathrm{p})\left(1-\mathrm{p}_{1}\right)\left(1-\mathrm{p}_{2}\right)}$
Here it has been assumed that the statement given by all the independent witnesses can be given in two ways only, so that if all the witnesses tell falsehoods they agree in telling the same falsehood. If this is not the case and c is the chance of their coincidence testimony then the Pr. that the statement is true = P p₁ p₂ Pr. that the statement is false = (1−p).c (1−p₁)(1−p₂)
However, chance of coincidence testimony is taken only if the joint statement is not contradicted by any witness.

Result − 7

A Probability Distribution spells out how a total probability of 1 is distributed over several values of a random variable.
Mean of any probability distribution of a random variable is given by
$\mu=\frac{\sum \mathrm{p}_{\mathrm{i}} \mathrm{x}_{\mathrm{i}}}{\sum \mathrm{p}_{\mathrm{i}}}=\sum \mathrm{p}_{\mathrm{i}} \mathrm{x}_{\mathrm{i}}$ ( Since Σ p_i = 1 )
Variance of a random variable is given by, σ² = ∑ ( x_i − µ)² . pi
σ² = ∑ p_i x²_i − µ² ( Note that SD = + $\sqrt{\sigma^{2}}$ )
The probability distribution for a binomial variate ‘X’ is given by; P ( X = r ) = ⁿC_rpⁿ q^n-r where all symbols have the same meaning as given in result 4. The recurrence formula
$\frac{\mathrm{P}(\mathrm{r}+1)}{\mathrm{P}(\mathrm{r})}=\frac{\mathrm{n}-\mathrm{r}}{\mathrm{r}+1} \cdot \frac{\mathrm{p}}{\mathrm{q}}$ is very helpful for quickly computing P(1) , P(2). P(3) etc. if P(0) is known.
Mean of BPD = np ; variance of BPD = npq.
If p represents a persons chance of success in any venture and ‘M’ the sum of money which he will receive in case of success, then his expectations or probable value = pM expectations = pM

Result − 8:
Geometrical Applications: The following statements are axiomatic:

If a point is taken at random on a given staright line AB, the chance that it falls on a particular segment PQ of the line is PQ/AB .
If a point is taken at random on the area S which includes an area σ, the chance that the point falls on σ is σ/S .

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